Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, if $\psi \in Aut(H)$, prove that $$N\rtimes_{\phi}H\cong N\rtimes_{\phi\circ\psi}H.$$

This is mentioned in the original post. I am seeking proof of the statement.

Edit: I have changed the order of $\phi$ and $\psi$ and the composition acts from right to left. Sorry about the vagueness.

Define $\varphi: N\rtimes_\phi H\to N\rtimes_{\phi\circ\psi}H$ by $$ (n,h)\mapsto (n,\psi^{-1}(h)) .$$ So it suffices to show that $\varphi$ is a group homomorphism.

\begin{align} \varphi((n_1,h_1)\cdot_\phi(n_2,h_2))&=\varphi((n_1\phi(h_1)(n_2),h_1h_2))\\ &=(n_1\phi(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2)) \end{align} On the other hand, \begin{align} \varphi((n_1,h_1))\cdot_{\phi\circ\psi}\varphi((n_2,h_2))&=(n_1,\psi^{-1}(h_1))\cdot_{\phi\circ\psi}(n_2,\psi^{-1}(h_2))\\ &=(n_1\phi\circ\psi\circ\psi^{-1}(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2)) \end{align}

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    $\begingroup$ Are you using the normal convention for composition? If so, the order of your composition is messed up. $\endgroup$ – Connor Malin May 14 '19 at 5:25
  • $\begingroup$ I undeleted it. I was confused because the order of your composition was incorrect which made my answer look wrong. Your composition right now is $H \rightarrow Aut(N)$ then $H \rightarrow H$ which doesn't make sense. $\endgroup$ – Connor Malin May 14 '19 at 5:32
  • $\begingroup$ Your definition of $\psi$ is that it is an automorphism of $H$. $\endgroup$ – Connor Malin May 14 '19 at 5:35
  • $\begingroup$ You must be using a different convention than I am. I am used to function composition being from right to left. $\endgroup$ – Connor Malin May 14 '19 at 6:09
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    $\begingroup$ @Connor I don't think there is a "normal convention for composition". There is the convention which you are brought up with (which depends on your upbringing), and some areas possibly favour one convention over the other. But neither is "normal" or "standard". $\endgroup$ – user1729 May 14 '19 at 8:45

$\DeclareMathOperator{\Aut}{Aut}$$\renewcommand{\phi}{\varphi}$I apologise for using a different notation (composition left-to-right and homo/auto-morphisms written as exponents), but if I do it differently, I might easily do it wrong. May try and translate it to different notation if deemed useful.

In $H \ltimes_{\phi} N$ the operation is given by $$ (h_{1}, n_{1}) \cdot (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2}). $$ (So in my notation $h_{2}^{\phi} \in \Aut(N)$ is the image of $h_{2}$ under $\phi$, and $n_{1}^{h_{2}^{\phi}}$ denotes its operation on $n_{1}$.)

In $H \ltimes_{\psi \phi} N$ the operation is given by $$ (h_{1}, n_{1}) \circ (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\psi \phi}} n_{2}). $$ (Here $\psi \phi$ is the composition of $\psi$ first and $\phi$ second.)

Now an isomorphism between the two groups is given by \begin{align} G :\ &H \ltimes_{\phi} N \to H \ltimes_{\psi \phi} N\\ &(h, n) \mapsto (h^{\psi^{-1}}, n). \end{align} (Note the critical inverse, see the calculation below.)

In fact

$$ G((h_{1}, n_{1})) \circ G((h_{2}, n_{2})) = (h_{1}^{\psi^{-1}}, n_{1}) \circ (h_{2}^{\psi^{-1}}, n_{2}) = (h_{1}^{\psi^{-1}} h_{2}^{\psi^{-1}}, n_{1}^{h_{2}^{\psi^{-1} \psi \phi}} n_{2}) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}), $$ which equals $$ G((h_{1}, n_{1}) \cdot (h_{2}, n_{2})) = G((h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2})) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}). $$


If you have two group structures on the same group, saying the identity is an isomorphism is saying the two group structures are actually the same. Here this couldn't possible be the case unless your automorphism is the identity.

Instead your map should be $(n,h) \rightarrow (n,\psi(h))$.


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